If Abc ~qrp ar(ABC)_9
ar(AQRP) 4
= AB = 18 cm and BC = 15 cm, then PR is
(a) 5 cm
(b) 10 cm (c) 15 cm (d) 20 cm
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option (b) 10 cm
as ratio of area of similar triangle is equal to ratio on square of there corresponding sides
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